Question

Bantu Jawab No 10 [Yang ada gambar mouse sampingnya] sama no 12

Answer 1

No.10
a.panjang CH = 17 cm
b.luas BCH =24 cm

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Answer 2

10.
CG = GH = BC = 12 cm

a. CH = ?
pakai cara pythagoras
CH(sisi miring) = √[(CG)^2 + (GH)^2]
= √(12^2 + 12^2)
= √(144 + 144)
= √288
= √(144 × 2)
= √12^2 × 2
= 12 √2 cm

b.
Luas segitiga BCH = BC × CH / 2
= 12 × 12 √2 / 2
= 12/2 × 12 √2
= 6 × 12 √2
= 72 √2 cm^2

12.
Luas segitiga STR = RT × TS / 2
24 = 8 × TS / 2
24 = 8/2 × TS
24 = 4TS
TS = 24/4 = 6 cm

Untuk mencari RS, pakai cara pythagoras
RS(sisi miring) = √[(RT)^2 + (TS)^2]
= √(6^2 + 8^2)
= √(36 + 64)
= √100
= √10^2
= 10 cm

RS = PS = 10 cm

PT = PS + ST = 10 + 6 = 16 cm

Luas trapesium PQRT = (PQ + RT) × PT / 2
= (18 + 8) × 16/2
= 26 × 8 = 208 cm^2

Luas segiempat PQRS = Luas trapesium PQRT - Luas segitiga STR
= 208 - 24 = 184 cm^2 (B)

^ = pangkat

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